Friday, November 23, 2012

Physics of Giant Balloons

As I was watching the Macy’s Thanksgiving Day Parade and all of the
giant balloons go by, I wondered just how hard it actually is to hold those
balloons down.

·Buoyant Force:

·The floats (filled with helium) are submerged in
the air surrounding them.

oFB
= PA = ρairghA = ρairgV = (ρV)g = mairg

oVolballoon
= 16,780 ft3 = 475 m3

oρair
= 1.225 kg/m3

o(1.225 kg/m3
* 475 m3)(9.8 m/s2) = 5,704 N

·Force of Gravity:

·We will assume that the nylon fabric the balloon
is made of does not contribute to the volume of the balloon, but for such a
large balloon, its mass is important.

oF = mballoong
= (muninflated + mHe)g

omuninflated
= 450 lbs = 204 kg

oρHe
= 0.1786 kg/m3

omHe
= ρV = 0.1786 kg/m3 * 475 m3 = 85 kg

omtot
= 204 kg + 85 kg = 289 kg

oF = 289 kg
* 9.8 m/s2 = 2,832 N

·Net Force of Balloon = FB – Fgrav

o5,704 N – 2,832
N = 2,872 N à2,800 N

·Will they be able to hold the Rugrats down?

·Each float is held down by 2 vehicles weighing
800 pounds, and a number of volunteers. The only requirement to be a balloon
handler is to weigh at least 120 pounds. The rugrats float requires a minimum
of 30 people to maneuver it.

o2 vehicles
(800 lbs) = 362.874 kg * 2 = 726 kg

o30 People
(at least 120 lbs) = 54.4311 kg * 30 = 1,633 kg

oTotal = 726
kg + 1,633 kg = 2,359 kg

o2359 kg *
9.8 m/s2 = 23,115 N à­– 23,000
N

·This will be more than enough force to keep the
balloon from floating away.

o2,800 N –
23,000 N = – 20,200 N

oWhy is all
of this force needed?

oTo maintain
the normal force so that the balloon handlers have enough friction to direct
the balloon.

o20,200 N / 23,000
N = 88% à the normal
force on each person on average is 88% what it would normally be. This means
that they only have 88% as much friction.